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  "body": "\n## Statement\n\nLet $f = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_0 \\in \\mathbb{Z}[x]$ with $n \\ge 1$ and $a_n \\neq 0$.\nLet $r = p/q$ be a rational [[root|00EHX6FWC0]] of $f$, with $p, q \\in \\mathbb{Z}$, $q \u003e 0$ and $\\gcd(p, q) = 1$.\nThen $p$ divides $a_0$ and $q$ divides $a_n$.\n\n## Remarks\n\nThe proof: multiply $f(p/q) = 0$ by $q^n$. Then $a_n p^n = -q(\\dots)$ and $a_0 q^n = -p(\\dots)$, and $\\gcd(p, q) = 1$ gives the two divisibilities.\nFor example, the only possible rational roots of $2x^3 - 3x + 1$ are $\\pm 1$ and $\\pm \\tfrac{1}{2}$. Evaluation shows that $1$ is a root.\nThe same statement holds over every [[unique factorisation domain|06YMXJMXYS]] and its field of fractions.\nFor a monic $f$, every rational root is an integer.\n\n## Explanations\n\n### L1 human\n\nThe theorem gives a finite list of candidates for the rational roots.\nYou test each candidate by evaluation.\nFor $x^3 - 2$, the candidates are $\\pm 1$ and $\\pm 2$, and none is a root, so $\\sqrt[3]{2}$ is irrational.\n",
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  "markdown": "https://math2.tamnd.com/node/0819CQ1V8H.md",
  "msc": [
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  "name": "Rational root theorem",
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